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  • hihocoder Arithmetic Expression【在线查询】

    Arithmetic Expression
     
    时间限制:2000ms
    单点时限:200ms
    内存限制:256MB

    描述

    Given N arithmetic expressions, can you tell whose result is closest to 9?

    输入

    Line 1: N (1 <= N <= 50000).
    Line 2..N+1: Each line contains an expression in the format of "a op b" where a, b are integers (-10000 <= a, b <= 10000) and op is one of addition (+), subtraction (-), multiplication (*) and division (/). There is no "divided by zero" expression.

    输出

    The index of expression whose result is closest to 9. If there are more than one such expressions, output the smallest index.

    样例输入
    4
    901 / 100
    3 * 3
    2 + 6
    8 - -1
    样例输出
    2

    【题意】:算出的结果最接近9的为第几个,相同的话输出靠前的。注意卡精度!要用double~
    【代码】:
    #include <bits/stdc++.h>
    #define LL long long
    #define maxn 500005
    const int inf = 0x3f3f3f3f;
    using namespace std;
    
    int n,idx;
    double a,b,sum=0;
    char op;
    double mi=0x3f3f3f3f3f;
    void cal()
    {
        switch(op)
        {
        case '+':
            sum=a+b;
            break;
        case '-':
            sum=a-b;
            break;
        case '*':
            sum=a*b;
            break;
        case '/':
            sum=a/b;
            break;
    
        }
    }
    int main()
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a>>op>>b;
            cal();
            double now = abs(9-sum);
            if(now < mi)//在线查询,边输入边查询,也可以叫打擂台算法,谁小谁上当min King,并记录是第几个人
            {
                idx=i;//而且是now < mi 不能等于,无形记录了字典序最小的那个人,因为后面出现的也是相等,不是小于了!
                mi=now;
            }
        }
        cout<<idx<<endl;
    }
    打擂台 严格小于
    
    
    
    
    
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  • 原文地址:https://www.cnblogs.com/Roni-i/p/7911245.html
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